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            <h1 id="seo-header">『算法-ACM竞赛-CodeForces』1245D-ShichikujiandPowerGri</h1>
            
            
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                <h1 id="『算法-ACM-竞赛-CodeForces』1245D-ShichikujiandPowerGri"><a href="#『算法-ACM-竞赛-CodeForces』1245D-ShichikujiandPowerGri" class="headerlink" title="『算法-ACM 竞赛-CodeForces』1245D-ShichikujiandPowerGri"></a>『算法-ACM 竞赛-CodeForces』1245D-ShichikujiandPowerGri</h1><h1 id="CodeForces-1245-D-Shichikuji-and-Power-Grid"><a href="#CodeForces-1245-D-Shichikuji-and-Power-Grid" class="headerlink" title="CodeForces - 1245 D - Shichikuji and Power Grid"></a>CodeForces - 1245 D - Shichikuji and Power Grid</h1><p>Shichikuji is the new resident deity of the South Black Snail Temple. Her first job is as follows:</p>
<p>There are nn new cities located in Prefecture X. Cities are numbered from 11 to nn. City ii is located xixi km North of the shrine and yiyi km East of the shrine. It is possible that (xi,yi)&#x3D;(xj,yj)(xi,yi)&#x3D;(xj,yj) even when i≠ji≠j.</p>
<p>Shichikuji must provide electricity to each city either by building a power station in that city, or by making a connection between that city and another one that already has electricity. So the City has electricity if it has a power station in it or it is connected to a City which has electricity by a direct connection or via a chain of connections.</p>
<ul>
<li>Building a power station in City ii will cost cici yen;</li>
<li>Making a connection between City ii and City jj will cost ki+kjki+kj yen per km of wire used for the connection. However, wires can only go the cardinal directions (North, South, East, West). Wires can cross each other. Each wire must have both of its endpoints in some cities. If City ii and City jj are connected by a wire, the wire will go through any shortest path from City ii to City jj. Thus, the length of the wire if City ii and City jj are connected is |xi−xj|+|yi−yj||xi−xj|+|yi−yj| km.</li>
</ul>
<p>Shichikuji wants to do this job spending as little money as possible, since according to her, there isn’t really anything else in the world other than money. However, she died when she was only in fifth grade so she is not smart enough for this. And thus, the new resident deity asks for your help.</p>
<p>And so, you have to provide Shichikuji with the following information: minimum amount of yen needed to provide electricity to all cities, the cities in which power stations will be built, and the connections to be made.</p>
<p>If there are multiple ways to choose the cities and the connections to obtain the construction of minimum price, then print any of them.</p>
<p>Input</p>
<p>First line of input contains a single integer nn (1≤n≤20001≤n≤2000) — the number of cities.</p>
<p>Then, nn lines follow. The ii-th line contains two space-separated integers xixi (1≤xi≤1061≤xi≤106) and yiyi (1≤yi≤1061≤yi≤106) — the coordinates of the ii-th city.</p>
<p>The next line contains nn space-separated integers c1,c2,…,cnc1,c2,…,cn (1≤ci≤1091≤ci≤109) — the cost of building a power station in the ii-th city.</p>
<p>The last line contains nn space-separated integers k1,k2,…,knk1,k2,…,kn (1≤ki≤1091≤ki≤109).</p>
<p>Output</p>
<p>In the first line print a single integer, denoting the minimum amount of yen needed.</p>
<p>Then, print an integer vv — the number of power stations to be built.</p>
<p>Next, print vv space-separated integers, denoting the indices of cities in which a power station will be built. Each number should be from 11 to nn and all numbers should be pairwise distinct. You can print the numbers in arbitrary order.</p>
<p>After that, print an integer ee — the number of connections to be made.</p>
<p>Finally, print ee pairs of integers aa and bb (1≤a,b≤n1≤a,b≤n, a≠ba≠b), denoting that a connection between City aa and City bb will be made. Each unordered pair of cities should be included at most once (for each (a,b)(a,b) there should be no more (a,b)(a,b) or (b,a)(b,a) pairs). You can print the pairs in arbitrary order.</p>
<p>If there are multiple ways to choose the cities and the connections to obtain the construction of minimum price, then print any of them.</p>
<p>Examples</p>
<p>Input</p>
<pre><code class="hljs">3
2 3
1 1
3 2
3 2 3
3 2 3
</code></pre>
<p>Output</p>
<pre><code class="hljs">8
3
1 2 3
0
</code></pre>
<p>Input</p>
<pre><code class="hljs">3
2 1
1 2
3 3
23 2 23
3 2 3
</code></pre>
<p>Output</p>
<pre><code class="hljs">27
1
2
2
1 2
2 3
</code></pre>
<p>Note</p>
<p>For the answers given in the samples, refer to the following diagrams (cities with power stations are colored green, other cities are colored blue, and wires are colored red):</p>
<p><img src="https://imgconvert.csdnimg.cn/aHR0cHM6Ly92ai56MTgwLmNuLzkzZTliZTk3YTlkYTdhMmUxMjE3OWQ5ODkzYmM5Mjdm?x-oss-process=image/format,png" srcset="/img/loading.gif" lazyload></p>
<p>For the first example, the cost of building power stations in all cities is 3+2+3&#x3D;83+2+3&#x3D;8. It can be shown that no configuration costs less than 8 yen.</p>
<p>For the second example, the cost of building a power station in City 2 is 2. The cost of connecting City 1 and City 2 is 2⋅(3+2)&#x3D;102⋅(3+2)&#x3D;10. The cost of connecting City 2 and City 3 is 3⋅(2+3)&#x3D;153⋅(2+3)&#x3D;15. Thus the total cost is 2+10+15&#x3D;272+10+15&#x3D;27. It can be shown that no configuration costs less than 27 yen.</p>
<p>这个题，先想朴素的算法，就是曼哈顿距离最小生成树，但是至少建一个电站，然后要做多少次最小生成树，对于三个点，建一个的时候有 3 种情况，两个电站 3 种，3 电站 3 种，然后情况挺多的，点这么多，那么这样必然超时，我们建立超级源点，源点到每个点的距离设置为每个点建电站的权值，然后一遍最小生成树即可，然后把边输出完事了，如果权值等于每个点建电站的权值和，那就是没有边，反正任意一种即可。然后完了！</p>
<pre><code class="hljs">#include &lt;bits/stdc++.h&gt;
using namespace std;
const int MAXN = 5000000 + 10;
const int MAXV = 2000 + 10;
typedef long long ll;
struct edge
&#123;
    int u, v;
    ll w;// 起点，终点，权值
&#125; e[MAXN];
struct node
&#123;
    int x, y;
&#125; t[MAXV], r[MAXV];
ll c[MAXV], k[MAXV];
int cont;
bool cmp(edge a, edge b)
&#123;
    return a.w &lt; b.w;
&#125;
int father[MAXV], n, cnt, dd, p[MAXV];;
ll ans;
int find(int x)
&#123;
    return father[x] == x ? x : father[x] = find(father[x]);
&#125;
int main()
&#123;
    scanf(&quot;%d&quot;, &amp;n);
    cont = 0;
    dd = 0;
    for (int i = 1; i &lt;= n; i++)
        father[i] = i;
    for (int i = 1; i &lt;= n; i++)
    &#123;
        scanf(&quot;%d %d&quot;, &amp;t[i].x, &amp;t[i].y);
    &#125;
    for (int i = 1; i &lt;= n; i++)
        scanf(&quot;%lld&quot;, &amp;c[i]);
    for (int i = 1; i &lt;= n; i++)
        scanf(&quot;%lld&quot;, &amp;k[i]);
    for (int i = 1; i &lt;= n; i++)
    &#123;
        for (int j = 1; j &lt; i; j++)
        &#123;
            e[++dd].u = i;
            e[dd].v = j;
            e[dd].w = (abs(t[i].x - t[j].x) + abs(t[i].y - t[j].y)) * (k[i] + k[j]);
        &#125;
    &#125;
    for (int i = 1; i &lt;= n; i++)
    &#123;
        e[++dd].u = 0;
        e[dd].v = i;
        e[dd].w = c[i];
    &#125;
    sort(e + 1, e + 1 + dd, cmp);
    for (int i = 1; i &lt;= dd; i++)
    &#123;
        int u = e[i].u, v = e[i].v;
        ll w = e[i].w;
        int fu = find(u), fv = find(v);
        if(fu == fv)
            continue;
        ++cont;
        r[cont].x = u;
        r[cont].y = v;
        ans += w;
        father[fv] = fu;
        cnt++;
        if(cnt == n)
            break;
    &#125;
    printf(&quot;%lld\n&quot;, ans);
    ans = 0;
    for (int i = 1; i &lt;= cont; i++)
    &#123;
        if(r[i].x == 0 || r[i].y == 0)
        &#123;
            ans++;
            p[ans] = r[i].x + r[i].y;
        &#125;
    &#125;
    printf(&quot;%lld\n&quot;, ans);
    for (int i = 1; i &lt;= ans; i++)
        printf(&quot;%d &quot;, p[i]);
    puts(&quot;&quot;);
    if(ans == cont)
        return puts(&quot;0&quot;),0;
    printf(&quot;%lld\n&quot;, cont - ans);
    for (int i = 1; i &lt;= cont; i++)
    &#123;
        if(r[i].x == 0 || r[i].y == 0)
            continue;
        printf(&quot;%d %d\n&quot;, r[i].x, r[i].y);
    &#125;
    return 0;
&#125;
</code></pre>

                
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